Lecture 6 : Aircraft orientation in 3 dimensions
Or describing simultaneous roll, pitch and yaw
G. Leng, Flight Dynamics, Stability & Control
1.0 Flight Dynamics Model
• For flight dynamics & control, the reference frame is
aligned with the aircraft and moves with it. (Why?)
Xb
Yb
Zb
Question: Where’s the origin located
for this body axes ?
G. Leng, Flight Dynamics, Stability & Control
• The aircraft is modelled as a rigid body with __ degrees of
freedom
• The ___ DOFs correspond to
• Denote the translational velocity of the aircraft by V = {u,v,w}
• Denote the angular velocity of the aircraft by  = {p,q,r}
G. Leng, Flight Dynamics, Stability & Control
Figure 1.1 : Six Degrees of Freedom
u
Yb
p - roll rate
v
Xb
q - pitch rate
r - yaw rate
Note the right hand
rule for rotation rates
Zb
w
G. Leng, Flight Dynamics, Stability & Control
1.1 Defining the aircraft orientation - Euler angles
The local horizon axes is aligned with the Earth fixed axes but
translated to the aircraft’s cg.
i
body axes
I
j
I
local horizon axes
k
K
J
Earth fixed axes
J
K
G. Leng, Flight Dynamics, Stability & Control
Question : How do we express the basis vectors IJK of the
local horizon system in terms of the basis vectors ijk of the
body axes system or vice versa ?
i
I
j
k
K
J
G. Leng, Flight Dynamics, Stability & Control
The local horizon axes system IJK can be rotated to coincide with
the body axes i j k by using three rotation angles or Euler angles
Yaw Euler angle

(Greek : psi)
Pitch Euler angle

(theta)
Roll Euler angle

(phi)
NB : The order of the rotations is important !
G. Leng, Flight Dynamics, Stability & Control
Step 1) Rotate IJK by an angle  about the K axis (yaw)
This yields the intermediate axes i1 j1 k1
J
i1
j1
I
K = k1
K
G. Leng, Flight Dynamics, Stability & Control
Step 1) Yaw about the K axis
i1
j1
k1
=
[ R ]
I

i1
I
J
K
J
j1
=
cos sin 0
-sin cos 0
0
0
1
I
J
K
G. Leng, Flight Dynamics, Stability & Control
Step 2) Rotate i1 j1 k1 by an angle  about the j1 axis (pitch)
This yields the intermediate axes i2 j2 k2
i2
i1
j1
j1 = j2
i1
k2
k1
k1
G. Leng, Flight Dynamics, Stability & Control
Step 2) Pitch about the j1 axis
i2

i2
j2
k2
=
[ R ]
i1
j1
k1
i1
k2
=
cos 0 -sin
0
1 0
sin 0 cos
i1
j1
k1
G. Leng, Flight Dynamics, Stability & Control
k1
Step 3) Rotate i2 j2 k2 by an angle  about the new i2 axis (roll)
This yields the body axes i j k
i2
i2 = i
j2
j2
j
k2
k
k2
G. Leng, Flight Dynamics, Stability & Control
Step 3) Roll about the i2 axis

i
j
k
=
[ R ]
i2
j2
k2
k
j
k2
=
1 0
0
0 cos sin
0 -sin cos
i2
j2
k2
G. Leng, Flight Dynamics, Stability & Control
j2
So finally, the basis vectors of the local horizon IJK and the
body axes ijk are related as follows :
i
j
k
=
[ R  ] [ R ] [ R ]
I
J
K
Question : What happens if we want IJK in terms of ijk ?
Hint : The rotation matrices have a special property
[ R ]-1
=
[ R ]T
G. Leng, Flight Dynamics, Stability & Control
Inverting the matrix product to get IJK in terms of ijk :
I
J
K
-1
=
=
i
j
k
[ R  ] [ R ] [ R ]
[R
]T
[ R
]T
[ R
]T
G. Leng, Flight Dynamics, Stability & Control
i
j
k
Exercise : Express the weight force component in terms of the
body axes basis vectors i j k
Hint : The weight component points downwards ie W K
hence we need only express K in terms of i j k
G. Leng, Flight Dynamics, Stability & Control
Using the relationship derived :
I
J
K
cos -sin 0
sin cos 0
0
0
1
=
cos 0 sin
0
1 0
-sin 0 cos
1 0
0
0 cos -sin
0 sin cos
i
j
k
Premultiply both sides by {0 0 1}
K
=
=
{-sin 0 cos } 1 0
0
0 cos -sin
0 sin cos
{-sin
cos sin
i
j
k
cos cos } i
j
k
G. Leng, Flight Dynamics, Stability & Control
What
happens if
the Euler
angles are
small ?
2.0 Rotation rates and change in aircraft orientation
How are the body rotational rates p,q,r related to the rate of
change of the Euler angles    ?
How about
p
=
d / dt
q
=
d / dt
r
=
d / dt
G. Leng, Flight Dynamics, Stability & Control
The angular velocity vector is  = p i + q j + r k written
using the body axes basis vectors
 describes the rate of change in orientation which can also
be written as :

’ K
=
+
’ j1 +
’ i2
Noting that
K
=
-sin i + cos sin j +
j1
=
j2
i2
=
i
=
cos cos k
cos j - sin k
G. Leng, Flight Dynamics, Stability & Control
p
q
r
-sin
=
0
1
cos
0
cos cos -sin
0
cos sin
’
’
’
Inverting yields :
'
'
'
=
=
=
p + (q sin + r cos ) tan
q cos - r sin
(q sin + r cos ) sec
What
happens if
 is 90 o ?
NB : These are the EOMs relating the rate of change of
aircraft orientation to body rotational rates
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Ex : Rotation rates during a sustained level turn
1. What is the angular velocity
vector for an aircraft executing a
level sustained turn of radius R
at speed V ? The bank angle is
.

2. Hence state the body axes
rotation rates.
Are they all zero ?
j
k
G. Leng, Flight Dynamics, Stability & Control
1. The turn rate =
V/R
2. Level turn  the rotation axis is vertical
3. Hence the angular velocity vector is

=
V/R {0, 0, 1}
=
V/R K
What is ambiguous ?
not the body axes k !
So how do we write this in body axes ?
G. Leng, Flight Dynamics, Stability & Control
4. Recall the relationship derived via Euler angles
K
=
-sin i + cos sin j + cos cos k
Hence in body axes

=
V/R(-sin i + cos sin j + cos cos k)
G. Leng, Flight Dynamics, Stability & Control
5. Recall  may be written in rotation rates as :

=
pi
+
qj
+
rk
Hence the body axes rotational rates for a sustained level turn are :
p
=
- V/R sin
q
=
V/Rcos sin
r
=
V/Rcos cos
What is the implication ?
G. Leng, Flight Dynamics, Stability & Control
Interpret this
6. What happens if we substitute these sustained, level turn
rotation rates
p
q
r
=
=
=
- V/R sin
V/Rcos sin
V/Rcos cos
in the equations for change in orientation ?
'
'
'
=
=
=
p + (q sin + r cos ) tan
q cos - r sin
(q sin + r cos ) sec
Physically what do you expect?
G. Leng, Flight Dynamics, Stability & Control
Follow-up exercise : Determine the rotation rates for a loop
G. Leng, Flight Dynamics, Stability & Control
3.0 Aerodynamic forces in the body axes
For a general aircraft orientation, the angle of attack a and
sideslip b are defined as follows:
Yb
a
b
Xb
Zb
V direction of flight
G. Leng, Flight Dynamics, Stability & Control
3.1 : Writing body velocity components with aerodynamic angles
Resolving the velocity vector
Yb
a
Xb
b
Zb
u
=
VT cos b cosa
v
=
VT sinb
w
=
VT cosb sina
VT
Question : What about the aerodynamic forces ?
G. Leng, Flight Dynamics, Stability & Control
3.2 Writing aerodynamic forces in the body axes
Often aerodynamic forces are specified in terms of 3 mutually
perpendicular forces
L
D
D : drag, aero force opposite to VT
S
L : lift, aero force perpendicular to VT
S : side force
a
Y
NB : L, D & S defines an axes system
i.e. the “flight path axes”
G. Leng, Flight Dynamics, Stability & Control
Z
b
VT
X
The transformation can be written in terms of two rotations
1) a rotation about the Y body axes by -a
2) a rotation about the resulting Z axis by b
and then inverting the components
D
S
L
cos(b) sin(b) 0
=
-
-sin(b) cos(b) 0
0
0
1
cos(-a) 0 -sin(-a)
0
1
0
sin(-a) 0 cos(-a)
G. Leng, Flight Dynamics, Stability & Control
X
Y
Z
Multiplying the rotation matrices yields the body axes components
for the aerodynamic forces
D
=
- (X cosa + Z sina) cosb - Y sinb
S
=
(X cosa + Z sina) sinb - Y cosb
L
=
X sina - Z cosa
G. Leng, Flight Dynamics, Stability & Control
or the inverse relation
X
=
L sina
+
(S sinb - D cosb) cosa
Y
=
- (S cosb
+
D sinb)
Z
=
- L cosa
+
(S sinb - D cosb) sina
Question : What do you expect if a and b are small ?
G. Leng, Flight Dynamics, Stability & Control